I am trying to work out the steps of the proof of the expression: $$\sum_n (\mathcal{E_n}-\mathcal{E_s})|\langle n|e^{i\mathbf{q}\cdot\mathbf{r}}|s \rangle|^2 = \frac{\hbar^2q^2}{2m}$$ from Eq. (5.48) in the book Principles of the Theory of Solids by Ziman. In the book it is mentioned that this can be shown by expanding out: $$[[\mathcal{H},e^{i\mathbf{q}\cdot\mathbf{r}}],e^{-i\mathbf{q}\cdot\mathbf{r}}]$$ where $$\mathcal{H} = -\frac{\hbar^2}{2m}\nabla^2+\mathcal{V}(\mathbf{r})$$ with $\mathcal{V}(\mathbf{r})$ being periodic. What I did is a simple expansion: $$[[\mathcal{H},e^{i\mathbf{q}\cdot\mathbf{r}}],e^{-i\mathbf{q}\cdot\mathbf{r}}] = 2\mathcal{H}-e^{i\mathbf{q}\cdot\mathbf{r}}\mathcal{H}e^{-i\mathbf{q}\cdot\mathbf{r}}-e^{-i\mathbf{q}\cdot\mathbf{r}}\mathcal{H}e^{i\mathbf{q}\cdot\mathbf{r}}$$ Then I took the inner product with the eigenstates of the Hamiltonian $\mathcal{H}|s\rangle = E_s |s\rangle$ to get $$\langle s|[[\mathcal{H},e^{i\mathbf{q}\cdot\mathbf{r}}],e^{-i\mathbf{q}\cdot\mathbf{r}}]|s\rangle = 2E_s - \langle s|e^{i\mathbf{q}\cdot\mathbf{r}}\mathcal{H}e^{-i\mathbf{q}\cdot\mathbf{r}}|s\rangle - \langle s|e^{-i\mathbf{q}\cdot\mathbf{r}}\mathcal{H}e^{i\mathbf{q}\cdot\mathbf{r}}|s\rangle$$ Now, what's obstructing my calculation is the fact that I cannot justify the last two terms in the above expression being equal. I really need them to be equal to show the top identity (also known as the Bethe sum rule). The main obstacle is the fact that $e^{i\mathbf{q}\cdot\mathbf{r}}$ is non-Hermitian.

I have found this identity in many books and journal articles. But I cannot find a satisfactory proof anywhere. One such example is this article:

Th result is stated in Eq. (3) and proof is given in section III. A. At the end of the proof they set $F = e^{i.\mathbf{q}.\mathbf{r}}$ to recover Eq. (3). The fact that they did not assume any form for $F$ means it must hold for any function. There is, however, one step that I cannot justify. In Eq. (9) how can they write: $$\langle 0|F(x)|l \rangle = \langle l|F(x)|0 \rangle$$ If I can even justify the above equality (for my case) then I'm set.

3 Answers
3

Start with the expression
$$\sum_n (\mathcal{E_n}-\mathcal{E_s})|\langle n|e^{i\mathbf{q}\cdot\mathbf{r}}|s \rangle|^2 =
\sum_n (\mathcal{E_n}-\mathcal{E_s})\langle s|e^{-i\mathbf{q}\cdot\mathbf{r}}|n \rangle\langle n|e^{i\mathbf{q}\cdot\mathbf{r}}|s \rangle. $$
The first trick is to realize that $(\mathcal{E_n}-\mathcal{E_s})e^{i\mathbf{q}\cdot\mathbf{r}}$ equals $\pm[\mathcal{H},e^{i\mathbf{q}\cdot\mathbf{r}}]$ when it is inside either of the two brackets:
$$\sum_n (\mathcal{E_n}-\mathcal{E_s})|\langle n|e^{i\mathbf{q}\cdot\mathbf{r}}|s \rangle|^2
=
\sum_n \langle s|e^{-i\mathbf{q}\cdot\mathbf{r}}|n \rangle\langle n|[\mathcal{H},e^{i\mathbf{q}\cdot\mathbf{r}}]|s \rangle
=
\langle s|e^{-i\mathbf{q}\cdot\mathbf{r}}[\mathcal{H},e^{i\mathbf{q}\cdot\mathbf{r}}]|s \rangle, $$
by summing $|n \rangle\langle n|$ to $1$. Analogously, you can do this on the first factor to get
$$\sum_n (\mathcal{E_n}-\mathcal{E_s})|\langle n|e^{i\mathbf{q}\cdot\mathbf{r}}|s \rangle|^2
=-
\langle s|[\mathcal{H},e^{-i\mathbf{q}\cdot\mathbf{r}}]e^{i\mathbf{q}\cdot\mathbf{r}}|s \rangle. $$

You now need to calculate the commutator. Since $\mathcal{V}(\mathbf{r})$ commutes with $e^{\pm i\mathbf{q}\cdot\mathbf{r}}$, you only need to worry about the kinetic term. Thus
$$
[\mathcal{H},e^{\pm i\mathbf{q}\cdot\mathbf{r}}]=\frac{1}{2m}[\mathbf{p}^2,e^{\pm i\mathbf{q}\cdot\mathbf{r}}]
=\frac{1}{2m}\mathbf{p}\cdot[\mathbf{p},e^{\pm i\mathbf{q}\cdot\mathbf{r}}]
+\frac{1}{2m}[\mathbf{p},e^{\pm i\mathbf{q}\cdot\mathbf{r}}]\cdot\mathbf{p},
$$
so
$$
[\mathcal{H},e^{\pm i\mathbf{q}\cdot\mathbf{r}}]=
\pm\frac{1}{2m}\mathbf{p}\cdot(\hbar \mathbf{q}e^{\pm i\mathbf{q}\cdot\mathbf{r}})
\pm\frac{1}{2m}(\hbar \mathbf{q}e^{\pm i\mathbf{q}\cdot\mathbf{r}})\cdot\mathbf{p}
=\pm\frac{\hbar}{2m}\mathbf{q}\cdot\left(e^{\pm i\mathbf{q}\cdot\mathbf{r}}\mathbf{p}+\mathbf{p}e^{\pm i\mathbf{q}\cdot\mathbf{r}}\right).
$$
To phrase this just right, you need to have $e^{+i\mathbf{q}\cdot\mathbf{r}}$ to the left or $e^{- i\mathbf{q}\cdot\mathbf{r}}$ to the right. Using the commutator $[\mathbf{p},e^{\pm i\mathbf{q}\cdot\mathbf{r}}]=\pm \hbar \mathbf{q}e^{\pm i\mathbf{q}\cdot\mathbf{r}}$, as above, you get
$$
[\mathcal{H},e^{+ i\mathbf{q}\cdot\mathbf{r}}]=
+\frac{\hbar}{2m}\mathbf{q}\cdot\left(e^{+ i\mathbf{q}\cdot\mathbf{r}}\mathbf{p}+\mathbf{p}e^{+ i\mathbf{q}\cdot\mathbf{r}}\right)
=
\frac{\hbar}{2m}e^{+ i\mathbf{q}\cdot\mathbf{r}}\mathbf{q}\cdot\left(2\mathbf{p}+\hbar\mathbf{q}\right)
$$
and
$$
[\mathcal{H},e^{- i\mathbf{q}\cdot\mathbf{r}}]=
-\frac{\hbar}{2m}\mathbf{q}\cdot\left(e^{- i\mathbf{q}\cdot\mathbf{r}}\mathbf{p}+\mathbf{p}e^{- i\mathbf{q}\cdot\mathbf{r}}\right)
=-
\frac{\hbar}{2m}\mathbf{q}\cdot\left(2\mathbf{p}+\hbar\mathbf{q}\right)e^{- i\mathbf{q}\cdot\mathbf{r}}.
$$

Putting either of these into the corresponding formula above, you get
$$\sum_n (\mathcal{E_n}-\mathcal{E_s})|\langle n|e^{i\mathbf{q}\cdot\mathbf{r}}|s \rangle|^2
=
\frac{\hbar}{2m}\langle s|\mathbf{q}\cdot\left(2\mathbf{p}+\hbar\mathbf{q}\right)|s \rangle
=
\frac{\hbar^2}{2m}\mathbf{q}^2+\frac{\hbar}{m}\langle s|\mathbf{q}\cdot\mathbf{p}|s \rangle. $$
You can then always force the mean value of $\mathbf{p}$ to vanish.

does not depend explicit on time. In particular, the point spectrum of energy eigenvalues $E_n$ does not depend on time. Let $\psi_n({\bf r})$ be the corresponding energy eigen-functions in position space, which satisfies the TISE

$$\tag{2} \hat{H}\psi_n({\bf r})~=~E_n \psi_n({\bf r}).$$

We may assume that $\psi_n({\bf r})\in{\mathbb{R}}$ is a real function, see also this Phys.SE post.

Let us consider a fixed time $t_0$. Let us pick a basis of time-dependent solutions

Yes, I know. I do in fact use the completeness relation as well as the fact that $\langle s|e^{i\mathbf{q}.\mathbf{r}}|n \rangle = \langle n|e^{-i\mathbf{q}.\mathbf{r}}|s \rangle^*$. That is not what is causing the problem. The problem arises from the fact that $|\langle s|e^{i\mathbf{q}.\mathbf{r}}|n \rangle|^2 \ne |\langle s|e^{-i\mathbf{q}.\mathbf{r}}|n \rangle|^2$ in general.
–
PhHEPDec 2 '12 at 12:37

Your second step (second last line) does not make any sense. You just pulled $e^{-i\mathbf{q}.\mathbf{r}}$ out of $\langle n|e^{-i\mathbf{q}.\mathbf{r}}| s \rangle$ and made it pass the $|n\rangle\langle n|$
–
PhHEPDec 6 '12 at 3:29

Sorry, that wasn't clear. $E_n$ commutes with $e^{i q \cdot r} $. So I just put it beside the eigenstate $\left|n \right>$ and then re-wrote $E_n\left|n \right>$ as $H\left|n \right>$. I also pulled out the $\left<s \right|$ on the left and $\left|s \right>$ on the right by linearity.
–
Stackexchange_user23Dec 6 '12 at 3:37

Okay, I see the missing steps. However, you are making use of $[H,e^{-i\mathbf{q}.\mathbf{r}}] = \hbar^2q^2/2m$. This is incorrect. The correct one is: $[[H,e^{-i\mathbf{q}.\mathbf{r}}],e^{i\mathbf{q}.\mathbf{r}}] = \hbar^2q^2/2m$. I can't believe I didn't notice this before
–
PhHEPDec 6 '12 at 6:07